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This common usage of % to mean m/v in biology is because of many biological solutions being dilute and water-based, an aqueous solution. Liquid water has a density of approximately 1 g/cm 3 (1 g/mL). Thus 100 mL of water is equal to approximately 100 g. Therefore, a solution with 1 g of solute dissolved in final volume of 100 mL aqueous ...
Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution. To dilute a solution means to add more solvent without the addition of more solute. The resulting solution is thoroughly mixed so as to ensure that all parts of the solution are ...
A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). This is incorrect because the unit "%" can only be used for dimensionless quantities.
The "dilution factor" is an expression which describes the ratio of the aliquot volume to the final volume. Dilution factor is a notation often used in commercial assays. For example, in solution with a 1/5 dilution factor (which may be abbreviated as x5 dilution ), entails combining 1 unit volume of solute (the material to be diluted) with ...
However, the names of all SI mass units are based on gram, rather than on kilogram; thus 10 3 kg is a megagram (10 6 g), not a *kilokilogram. The tonne (t) is an SI-compatible unit of mass equal to a megagram (Mg), or 10 3 kg. The unit is in common use for masses above about 10 3 kg and is often used with SI prefixes.
Normality is defined as the number of gram or mole equivalents of solute present in one liter of solution.The SI unit of normality is equivalents per liter (Eq/L). = where N is normality, m sol is the mass of solute in grams, EW sol is the equivalent weight of solute, and V soln is the volume of the entire solution in liters.
An aqueous solution containing 2 g of glucose and 2 g of fructose per 100 g of solution contains 2/100=2% glucose on a wet basis, but 2/4=50% glucose on a dry basis.If the solution had contained 2 g of glucose and 3 g of fructose, it would still have contained 2% glucose on a wet basis, but only 2/5=40% glucose on a dry basis.
The quantity "1 ppm" can be used for a mass fraction if a water-borne pollutant is present at one-millionth of a gram per gram of sample solution. When working with aqueous solutions, it is common to assume that the density of water is 1.00 g/mL. Therefore, it is common to equate 1 kilogram of water with 1 L of water.